I have to give an estimation of the value of this series
$$\sum_{n = 0}^{+\infty} \dfrac{3^n}{n!}$$
I know this series does converge to $e^3$, but I am not allowed to use this fact. Then here is what I thought of:
$$\dfrac{3^n}{n!} = \dfrac{\overbrace{3\cdot 3\cdot 3\cdot 3\cdots 3}^{n \ \text{times}}}{1\cdot 2\cdot 3\cdots n} = \dfrac{3\cdot 3\cdot 3}{1\cdot 2\cdot 3}\cdot \dfrac{3\cdots 3}{\cdots n} = \dfrac{9}{2} \cdot \dfrac{3\cdot 3\cdots 3}{4\cdot 5\cdots n}$$
Now I majorized: $n! \geq 4^{n-3}$ hence $\frac{1}{n!} \leq \frac{1}{4^{n-3}}$ so then
$$\dfrac{3^n}{n!} \leq \dfrac{9}{2}\cdot \dfrac{3^{n-3}}{4^{n-3}}$$
So then
$$\sum_{n = 0}^{+\infty} \dfrac{3^n}{n!} \leq \dfrac{9}{2}\cdot \dfrac{64}{27}\sum_{n = 3}^{+\infty} \left(\dfrac{3}{4}\right)^n$$
Using the geometric progression and substracting the exceeding terms:
$$\sum_{n= 3}^{+\infty} = \sum_{n = 0}^{+\infty} -\ \text{term of n = 0}\ - \ \text{term of n = 1}\ - \ \text{term of n = 2}$$
I got
$$\sum_{n = 0}^{+\infty} \dfrac{3^n}{n!} \leq 18$$
Now this is wrong for $e^3 \approx 20.08$. Where did I went wrong?
You have $$\sum_{n = {\color{red}3}}^{+\infty} \dfrac{3^n}{n!} \leq \dfrac{9}{2}\cdot \dfrac{64}{27}\sum_{n = 3}^{+\infty} \left(\dfrac{3}{4}\right)^n=18$$ and so $$\sum_{n = {\color{red}0}}^{+\infty} \dfrac{3^n}{n!} \leq \frac{17}2+18=26.5.$$